Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;
#define INF 10000000000000
struct edge
{
    long long to, cost;
};
struct graph
{
    int V;                  //頂点数
    vector<vector<edge>> G; //頂点sからの最短距離
    vector<long long> d;
    graph(int n)
    {
        init(n);
    }
    void init(int n)
    {
        V = n;
        G.resize(V);
        d.resize(V);
        for (int i = 0; i < V; i++)
        {
            d[i] = INF;
        }
    }
    void add_edge(int s, int t, long long cost)
    {
        edge e;
        e.to = t;      //行き先
        e.cost = cost; //cost
        G[s].push_back(e);
    }
    void dijkstra(int s)
    {
        for (int i = 0; i < V; i++)
        {
            d[i] = INF;
        }
        d[s] = 0;
        priority_queue<P, vector<P>, greater<P>> que;
        que.push(P(0, s));
        while (!que.empty())
        {
            P p = que.top();
            que.pop();
            long long v = p.second;
            if (d[v] < p.first)
                continue;
            for (auto e : G[v])
            {
                if (d[e.to] > d[v] + e.cost)
                {
                    d[e.to] = d[v] + e.cost;
                    que.push(P(d[e.to], e.to));
                }
            }
        }
    }
};

int main()
{
    int N, M, s, t;
    cin >> N >> M >> s >> t;
    s--, t--;
    graph G1(N);
    graph G2(N);
    for (int i = 0; i < M; i++)
    {
        int x, y;
        cin >> x >> y;
        x--, y--;
        G1.add_edge(x, y, 1);
        G2.add_edge(x, y, 1);
        G1.add_edge(y, x, 1);
        G2.add_edge(y, x, 1);
    }
    G1.dijkstra(s);
    G2.dijkstra(t);
    map<long long, long long> g1, g2;
    for (int i = 0; i < N; i++)
    {
        g1[G1.d[i]]++;
        g2[G2.d[i]]++;
    }
    long long ans = 0;
    int dis = G1.d[t];
    for (int i = 0; i <= dis - 2; i++)
    {
        ans += g1[i] * g2[dis - i - 2];
    }
    cout << ans << endl;
}